How to show that condition number of $QA$ is equal to condition number of $A$

Question

If $A$ is an invertible matrix and $Q$ is an orthonormal matrix, show that $$k(QA) = k(A).$$

Hint: $k(A) = \frac{\sigma_{1}}{\sigma_{n}}$ (the ratio of the largest and smallest eigenvalues).

Answer 1

using polar form, $A$ may be written uniquely as
$A=UP$ where $P$ is (real symmetric) Positive Definite. $P$ contains the singular values of $A$.

$\big(QA\big)=Q\big(A\big)=Q\big(UP\big) = \big(QU\big)P$
and $P$ has the singular values for $\big(QA\big)$ so $\big(QA\big)$ and A have the same singular values.

note: this assumes $Q$ is a real orthogonal matrix. There really isn't such thing as an 'orthonormal matrix'.